You want to generate points that are uniformly distributed over the interior of a 3D unit sphere -- meaning every sub-volume of equal size is equally likely to contain a point.
Describe two methods to do this: rejection sampling and the inverse CDF (direct) method. For each, explain the procedure and, for rejection sampling, compute the acceptance rate.
For a point $(X, Y, Z)$ drawn uniformly from the ball, derive the marginal PDF of a single coordinate -- say $X$. What does it look like, and where is the density concentrated?
Hints
For rejection sampling, think about what fraction of the bounding cube $[-1,1]^3$ is actually inside the unit ball -- that gives you the acceptance rate directly.
For the direct method, figure out how radius $R$ must be distributed: in 3D, volume scales as $r^3$, so the CDF of $R$ is $F_R(r) = r^3$. Invert this to sample $R$, and use a normalized Gaussian vector for the direction.
For the marginal of $X$: at a fixed $x$, the valid $(y, z)$ pairs fill a disk of radius $\sqrt{1 - x^2}$. The PDF $f_X(x)$ is proportional to the area of that disk, $\pi(1 - x^2)$, normalized over $[-1, 1]$.
Worked Solution
How to Think About It: Uniform sampling from a 3D ball trips people up because the naive approach fails: if you just pick $R \sim \text{Unif}(0,1)$ and a random direction, you over-sample near the center. The key issue is that volume grows as $r^2 \, dr$ in 3D -- there is far more volume near $r = 1$ than near $r = 0$. So $R$ must have a distribution that accounts for this volume weighting. Two practical methods: rejection sampling (simple to implement, decent efficiency) and the inverse CDF method (exact, no wasted samples). For the marginal, think about how many points with $x$-coordinate near some value $x$ exist -- you are integrating over a disk of radius $\sqrt{1 - x^2}$, so the density is proportional to the area of that disk.
Quick Estimate: Rejection sampling acceptance rate: we are drawing from a cube $[-1,1]^3$ (volume 8) and keeping points inside the unit ball (volume $\frac{4}{3}\pi$). Acceptance rate $= \frac{4\pi/3}{8} = \frac{\pi}{6} \approx 0.524$. So roughly half your samples are accepted -- quite efficient for 3D. In $d$ dimensions this degrades fast: at $d = 10$, the ball fills only about 0.25% of the bounding hypercube, making rejection sampling essentially useless.
Formal Solution:
Method 1 -- Rejection Sampling:
Draw $(x, y, z) \sim \text{Unif}[-1, 1]^3$ (three independent uniform draws on $[-1, 1]$).
Accept if $x^2 + y^2 + z^2 \leq 1$; otherwise reject and repeat.
The volume of the cube is
^3 = 8$. The volume of the unit ball is $\frac{4}{3}\pi$. Acceptance probability:
Conditional on acceptance, the distribution is exactly uniform over the ball by symmetry.
Method 2 -- Inverse CDF (Direct Method):
First, find the CDF of the radial coordinate $R$ for a uniform ball point. In 3D, volume scales as $r^3$, so:
$F_R(r) = r^3, \quad r \in [0, 1]$
Inverting: if $U \sim \text{Unif}(0, 1)$, set $R = U^{1/3}$.
For the direction, draw $(G_1, G_2, G_3) \stackrel{\text{iid}}{\sim} N(0, 1)$ and normalize to get a uniform point on the sphere surface. Then the final point is:
This produces exactly one accepted point per draw -- no waste.
Part 2 -- Marginal Distribution of $X$:
For a fixed value $x \in [-1, 1]$, the cross-section of the ball perpendicular to the $x$-axis is a disk of radius $\sqrt{1 - x^2}$ and area $\pi(1 - x^2)$. The PDF of $X$ is proportional to this cross-sectional area:
$\boxed{f_X(x) = \frac{3}{4}(1 - x^2), \quad x \in [-1, 1]}$
Answer:
Rejection acceptance rate: $\pi/6 \approx 52.4\%$
Direct method: $R = U^{1/3}$, direction from normalized Gaussian vector
Marginal PDF: $f_X(x) = \frac{3}{4}(1 - x^2)$, concentrated near $x = 0$, vanishing at $\pm 1$
Intuition
The core lesson here is that uniform over a ball is not the same as uniform in each coordinate independently, and it is not uniform in the radius either. Volume in 3D accumulates as $r^2 \, dr$, so the outer shell near $r = 1$ contains far more volume than the center -- which is why you must transform a uniform $U$ into $R = U^{1/3}$ rather than using $R = U$ directly. This is a specific case of the general principle: when sampling from a region in $d$ dimensions, you need to account for how the "measure" (volume, area) changes with each coordinate.
The marginal distribution result -- $f_X(x) = \frac{3}{4}(1 - x^2)$ -- shows up naturally when you think about cross-sections. It is not uniform, and it is not Beta in the standard sense, but it has the same qualitative shape: density peaks at the center ($x = 0$) and tapers off toward the edges. This matters in practice: if you are working with high-dimensional data on or near a sphere (common in machine learning and factor models), the projection of your data onto any single axis will look like this distribution, not a flat one. Understanding geometric distortion from high-dimensional spaces is directly relevant to portfolio risk modeling, factor decompositions, and random matrix theory.